Question 24050
You can set up the appropriate equation for this situation from the problem description.
Let x = the required number of quarts of pure (100%) antifreze, then (20-x) = the required number of quarts of 40% antifreeze solution which, when added together, will give 20 quarts of 50% antifreeze solution. Change the percentages to their decimal equivalents. The equation is:
{{{x + (20-x)(0.4) = 20(0.5)}}} Simplify and solve for x.
{{{x - 0.4x + 8 = 10}}}
{{{0.6x + 8 = 10}}} Subtract 8 from both sides of the equation.
{{{0.6x = 2}}} Divide both sides by 0.6
{{{x = 2/0.6}}} = 3 1/3
x = 3 1/3 quarts of pure antifreeze.
20-x = 16 2/3 quarts of 40% antifreeze.